Greedy Algorithms:

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Greedy Algorithms

• A greedy algorithm always makes the choice that
  looks best at the moment.
• It makes a local optimal choice in the hope that
  this choice will lead to a globally optimal
  solution.
• Greedy algorithms yield optimal solutions for
  many (but not all) problems.

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The 0-1 Knapsack problem

• The 0-1 knapsack problem
• N items, where the i-th item is worth vi dollars
  and weight wi pounds. 11 p 3
  p 4p 58 p 8p 88p
• vi and wi are integers.
• 
  3 6 35 8 28 66
• We can carry at most W (integer) pounds.
• How to take as valuable a load as possible.
• An item cannot be divided into pieces.
• The fractional knapsack problem
• The same setting, but the thief can take
  fractions of items.
• W may not be integer.

W
3
Solve the fractional Knapsack problem

• Greedy on the value per pound vi/wi.
• Each time, take the item with maximum vi/wi .
• If exceeds W, take fractions of the item.
• Example (1, 5), (2, 9), (3, 12), (3, 11)
  and w4.
• vi/wi 5 4.5 4.0
  3.667
• First (1, 5), Second (2, 9), Third
  1/3 (3, 12)
• Total W 1, 3,
  4.
• 
  
• 
  Can only take part
  of item

4
Proof of correctness (The hard part)
X i1 w1 i2 w2 wj . . . ip wp . . . ikwk

• Let X i1, i2, ik be the optimal items taken.
• Consider the item j (vj, wj) with the highest
  v /w.
• if j is not used in X (the optimal solution),
  get rid of some items with total weight wj
  (possibly fractional items) and add item j.
• (since fractional items are allowed, we can
  do it.)
• Total value is increased. Why?
• One more item selected by greedy is added to X
• Repeat the process, X is changed to contain all
  items selected by greedy WITHOUT decreasing the
  total value taken by the thief.

5
The 0-1 knapsack problem cannot be solved
optimally by greedy

• Counter example (moderate part)
• W10 2 1.8
• Items found (6pounds, 12dollars), (5pounds, 9
  dollar),
• 1.8 1.
  1
• (5pounds, 9 dollars), (3pounds, 3 dollars), (3
  pounds, 3 dollars)
• If we first take (6, 12) according to greedy
  algorithm, then solution is (6, 12), (3, 3)
  (total value is 12315).
• However, a better solution is (5, 9), (5, 9)
  with total value 18.
• To show that a statement does not hold, we only
  have to give an example.

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A subset of mutually compatibles jobs c, f
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Sorting the n jobs based on fi needs O(nlog n)
time
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Example

• Jobs (s, f) (0, 10), (3, 4), (2, 8), (1, 5),
• (4, 5), (4, 8), (5, 6) (7,9).
• Sorting based on fi
• (3, 4) (1, 5), (4, 5) (5, 6) (4,8) (2,8)
• (7, 9)(0,10).
• Selecting jobs
• (3,4) (4, 5) (5,6) (7, 9)

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Sort ob finish time b, c, a, e, d, f, g,
h. Greedy algorithm Selects b, e, h.
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Depth The maximum No. of jobs required at any
time.
Depth3
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Depth The maximum No. of jobs required at any
time.
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Greedy on start time
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Greedy Algorithm
c
d
g
j
b
f
i
Depth The maximum No. of jobs required at any
time.
a
e
h
Depth3
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d?depth
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l10, l21
l20, l10
0
11
10
l19, l20
l10, l21
1
11
0
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diltdj
We check every pair of consecutive numbers, if
there is not inversion, then the whole sequenc
has no inversion.
n1?n2? ?nk Example 1, 2, 3, 4, 5, 6, 7, 8, 9
1, 2, 6, 7, 3, 4, 5, 8, 9
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diltdj
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Example

• Job ti di
• 2 2
• 3 4
• 4 6
• 4 8
• 6 10

j1
5
9
13
19
j2
j5
2
j3
j4
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